Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Scanner;

public class KarenandGame {

	public static void main(String[] args) throws IOException {
		// TODO Auto-generated method stub

		//Scanner input = new Scanner(System.in);
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] in = br.readLine().split(" ");
		int n = Integer.parseInt(in[0]);
		int m = Integer.parseInt(in[1]);
		ArrayList<String> list =  new ArrayList<String>();
		ArrayList<String> list2 = new ArrayList<String>();
		int[][] arr = new int[n][m];
		for(int i=0;i<n;i++){
			in = br.readLine().split(" ");
			for(int j=0; j<m;j++){
				arr[i][j] = Integer.parseInt(in[j]);
			}
		}
		int[] min = new int[n];
		int[] min1  = new int[m];
		for(int i=0;i<n;i++){
			min[i] = 501;
			for(int j=0; j<m;j++){
				if(arr[i][j]<min[i])min[i] = arr[i][j];
			}
			for(int j=0;j<min[i];j++){
				list.add("row "+(i+1));
				
			}
		}
		for(int j=0;j<m;j++){
			min1[j] = 501;
			for(int i=0;i<n;i++){
				if(arr[i][j]<min1[j])min1[j] = arr[i][j];
			}
			for(int i=0; i<min1[j];i++){
				list2.add("col "+(j+1));
			}
		}
		for(int j=0; j<m;j++){
			int prev = arr[0][j]-min[0];
			for(int i=1;i<n;i++){
				if(arr[i][j]-min[i]!=prev){
					System.out.println(-1);
					return;
					
				}
			}
			for(int i=0;i<prev;i++){
				list.add("col "+(j+1));
			}
		}
		for(int i=0; i<n;i++){
			int prev = arr[i][0] - min1[0];
			for(int j=1;j<m;j++){
				if(arr[i][j]-min1[j]!=prev){
					System.out.println(-1);
					return;
				}
			}
			for(int j=0; j<prev;j++){
				list2.add("row "+(i+1));			}
		}
		if(list.size() < list2.size()){
			System.out.println(list.size());
			for(int i=0; i<list.size();i++)System.out.println(list.get(i));
			
		}
		else{
			System.out.println(list2.size());
			for(int i=0; i<list2.size();i++)System.out.println(list2.get(i));
		}
		
		
	}

}